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# Let we are given a series as ${{a}_{1}},{{a}_{2}},{{a}_{3}},....{{a}_{10}}$ be in GP with ${{a}_{1}}>0$ for I = 1, 2, …..10 and S be set of the pairs (r, k) such that $r,k\in N$ (The set of natural numbers) for which $\left| \begin{matrix} {{\log }_{e}}{{a}_{1}}^{r}{{a}_{2}}^{k} & {{\log }_{e}}{{a}_{2}}^{r}{{a}_{3}}^{k} & {{\log }_{e}}{{a}_{3}}^{r}{{a}_{4}}^{k} \\ {{\log }_{e}}{{a}_{4}}^{r}{{a}_{5}}^{k} & {{\log }_{e}}{{a}_{5}}^{r}{{a}_{6}}^{k} & {{\log }_{e}}{{a}_{6}}^{r}{{a}_{7}}^{k} \\ {{\log }_{e}}{{a}_{7}}^{r}{{a}_{8}}^{k} & {{\log }_{e}}{{a}_{8}}^{r}{{a}_{9}}^{k} & {{\log }_{e}}{{a}_{9}}^{r}{{a}_{10}}^{k} \\\end{matrix} \right|=0$\begin{align} & a)\text{ Infinitely many} \\ & \text{b) 4} \\ & \text{c) 10} \\ & \text{d) 2} \\ \end{align}

Last updated date: 13th Jun 2024
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Hint: Now to solve this consider first consider the given GP. Now since the given numbers ${{a}_{1}},{{a}_{2}},{{a}_{3}},....{{a}_{10}}$ are in GP, we have $\dfrac{{{a}_{2}}}{{{a}_{1}}}=\dfrac{{{a}_{3}}}{{{a}_{2}}}=....=\dfrac{{{a}_{10}}}{{{a}_{9}}}$ . Since the ratio of two consecutive terms taken in order in GP are the same. Now consider the given determinant. We will use column transformation ${{C}_{1}}\to {{C}_{1}}-{{C}_{2}}$ and ${{C}_{2}}={{C}_{2}}-{{C}_{3}}$ in the given determinant. Now we will use the properties of log which says $\log a-\log b=\log \left( \dfrac{a}{b} \right)$ . Now further simplifying the determinant we will make use of the equation $\dfrac{{{a}_{2}}}{{{a}_{1}}}=\dfrac{{{a}_{3}}}{{{a}_{2}}}=....=\dfrac{{{a}_{10}}}{{{a}_{9}}}$ and hence get a simplified determinant. Now we can easily find the answer.

Complete step-by-step solution
Now first let us consider the given terms.
The terms ${{a}_{1}},{{a}_{2}},{{a}_{3}},....{{a}_{10}}$ are given to be in GP. This means the ratio of consecutive terms taken in order is the same. Hence we can write.
$\dfrac{{{a}_{2}}}{{{a}_{1}}}=\dfrac{{{a}_{3}}}{{{a}_{2}}}=....=\dfrac{{{a}_{10}}}{{{a}_{9}}}$ . Let us call this common ratio as d.
Hence we get, $\dfrac{{{a}_{2}}}{{{a}_{1}}}=\dfrac{{{a}_{3}}}{{{a}_{2}}}=....=\dfrac{{{a}_{10}}}{{{a}_{9}}}=d$
Now from the above equation we can say that ${{a}_{n}}=d{{a}_{n-1}}....................\left( 1 \right)$
Now consider the given determinant. $\left| \begin{matrix} {{\log }_{e}}{{a}_{1}}^{r}{{a}_{2}}^{k} & {{\log }_{e}}{{a}_{2}}^{r}{{a}_{3}}^{k} & {{\log }_{e}}{{a}_{3}}^{r}{{a}_{4}}^{k} \\ {{\log }_{e}}{{a}_{4}}^{r}{{a}_{5}}^{k} & {{\log }_{e}}{{a}_{5}}^{r}{{a}_{6}}^{k} & {{\log }_{e}}{{a}_{6}}^{r}{{a}_{7}}^{k} \\ {{\log }_{e}}{{a}_{7}}^{r}{{a}_{8}}^{k} & {{\log }_{e}}{{a}_{8}}^{r}{{a}_{9}}^{k} & {{\log }_{e}}{{a}_{9}}^{r}{{a}_{10}}^{k} \\ \end{matrix} \right|$
Now to this determinant first we will apply column transformation ${{C}_{1}}\to {{C}_{1}}-{{C}_{2}}$
Hence we get $\left| \begin{matrix} {{\log }_{e}}{{a}_{1}}^{r}{{a}_{2}}^{k}-{{\log }_{e}}{{a}_{2}}^{r}{{a}_{3}}^{k} & {{\log }_{e}}{{a}_{2}}^{r}{{a}_{3}}^{k} & {{\log }_{e}}{{a}_{3}}^{r}{{a}_{4}}^{k} \\ {{\log }_{e}}{{a}_{4}}^{r}{{a}_{5}}^{k}-{{\log }_{e}}{{a}_{5}}^{r}{{a}_{6}}^{k} & {{\log }_{e}}{{a}_{5}}^{r}{{a}_{6}}^{k} & {{\log }_{e}}{{a}_{6}}^{r}{{a}_{7}}^{k} \\ {{\log }_{e}}{{a}_{7}}^{r}{{a}_{8}}^{k}-{{\log }_{e}}{{a}_{8}}^{r}{{a}_{9}}^{k} & {{\log }_{e}}{{a}_{8}}^{r}{{a}_{9}}^{k} & {{\log }_{e}}{{a}_{9}}^{r}{{a}_{10}}^{k} \\ \end{matrix} \right|$
Now again we will use a column transformation, ${{C}_{2}}={{C}_{2}}-{{C}_{3}}$ and hence we will get.
$\left| \begin{matrix} {{\log }_{e}}{{a}_{1}}^{r}{{a}_{2}}^{k}-{{\log }_{e}}{{a}_{2}}^{r}{{a}_{3}}^{k} & {{\log }_{e}}{{a}_{2}}^{r}{{a}_{3}}^{k}-{{\log }_{e}}{{a}_{3}}^{r}{{a}_{4}}^{k} & {{\log }_{e}}{{a}_{3}}^{r}{{a}_{4}}^{k} \\ {{\log }_{e}}{{a}_{4}}^{r}{{a}_{5}}^{k}-{{\log }_{e}}{{a}_{5}}^{r}{{a}_{6}}^{k} & {{\log }_{e}}{{a}_{5}}^{r}{{a}_{6}}^{k}-{{\log }_{e}}{{a}_{6}}^{r}{{a}_{7}}^{k} & {{\log }_{e}}{{a}_{6}}^{r}{{a}_{7}}^{k} \\ {{\log }_{e}}{{a}_{7}}^{r}{{a}_{8}}^{k}-{{\log }_{e}}{{a}_{8}}^{r}{{a}_{9}}^{k} & {{\log }_{e}}{{a}_{8}}^{r}{{a}_{9}}^{k}-{{\log }_{e}}{{a}_{9}}^{r}{{a}_{10}}^{k} & {{\log }_{e}}{{a}_{9}}^{r}{{a}_{10}}^{k} \\ \end{matrix} \right|$
Now we know that $\log a-\log b=\log \left( \dfrac{a}{b} \right)$ using this property we get.
$\left| \begin{matrix} {{\log }_{e}}\dfrac{{{a}_{1}}^{r}{{a}_{2}}^{k}}{{{a}_{2}}^{r}{{a}_{3}}^{k}} & {{\log }_{e}}\dfrac{{{a}_{2}}^{r}{{a}_{3}}^{k}}{{{a}_{3}}^{r}{{a}_{4}}^{k}} & {{\log }_{e}}{{a}_{3}}^{r}{{a}_{4}}^{k} \\ {{\log }_{e}}\dfrac{{{a}_{4}}^{r}{{a}_{5}}^{k}}{{{a}_{5}}^{r}{{a}_{6}}^{k}} & {{\log }_{e}}\dfrac{{{a}_{5}}^{r}{{a}_{6}}^{k}}{{{a}_{6}}^{r}{{a}_{7}}^{k}} & {{\log }_{e}}{{a}_{6}}^{r}{{a}_{7}}^{k} \\ {{\log }_{e}}\dfrac{{{a}_{7}}^{r}{{a}_{8}}^{k}}{{{a}_{8}}^{r}{{a}_{9}}^{k}} & {{\log }_{e}}\dfrac{{{a}_{8}}^{r}{{a}_{9}}^{k}}{{{a}_{9}}^{r}{{a}_{10}}^{k}} & {{\log }_{e}}{{a}_{9}}^{r}{{a}_{10}}^{k} \\ \end{matrix} \right|$
Now from equation (1) we get.
$\left| \begin{matrix} {{\log }_{e}}\dfrac{{{a}_{1}}^{r}{{\left( d{{a}_{1}} \right)}^{k}}}{{{a}_{2}}^{r}{{\left( d{{a}_{2}} \right)}^{k}}} & {{\log }_{e}}\dfrac{{{a}_{2}}^{r}{{\left( d{{a}_{2}} \right)}^{k}}}{{{a}_{3}}^{r}{{\left( d{{a}_{3}} \right)}^{k}}} & {{\log }_{e}}{{a}_{3}}^{r}{{a}_{4}}^{k} \\ {{\log }_{e}}\dfrac{{{a}_{4}}^{r}{{\left( d{{a}_{4}} \right)}^{k}}}{{{a}_{5}}^{r}{{\left( d{{a}_{5}} \right)}^{k}}} & {{\log }_{e}}\dfrac{{{a}_{5}}^{r}{{\left( d{{a}_{5}} \right)}^{k}}}{{{a}_{6}}^{r}{{\left( d{{a}_{6}} \right)}^{k}}} & {{\log }_{e}}{{a}_{6}}^{r}{{a}_{7}}^{k} \\ {{\log }_{e}}\dfrac{{{a}_{7}}^{r}{{\left( d{{a}_{7}} \right)}^{k}}}{{{a}_{8}}^{r}{{\left( d{{a}_{8}} \right)}^{k}}} & {{\log }_{e}}\dfrac{{{a}_{8}}^{r}{{\left( d{{a}_{8}} \right)}^{k}}}{{{a}_{9}}^{r}{{\left( d{{a}_{9}} \right)}^{k}}} & {{\log }_{e}}{{a}_{9}}^{r}{{a}_{10}}^{k} \\ \end{matrix} \right|$
Now simplifying by we get
$\left| \begin{matrix} {{\log }_{e}}\dfrac{{{a}_{1}}^{r+k}}{{{a}_{2}}^{r+k}} & {{\log }_{e}}\dfrac{{{a}_{2}}^{r+k}}{{{a}_{3}}^{r+k}} & {{\log }_{e}}{{a}_{3}}^{r}{{a}_{4}}^{k} \\ {{\log }_{e}}\dfrac{{{a}_{4}}^{r+k}}{{{a}_{5}}^{r+k}} & {{\log }_{e}}\dfrac{{{a}_{5}}^{r+k}}{{{a}_{6}}^{r+k}} & {{\log }_{e}}{{a}_{6}}^{r}{{a}_{7}}^{k} \\ {{\log }_{e}}\dfrac{{{a}_{7}}^{r+k}}{{{a}_{8}}^{r+k}} & {{\log }_{e}}\dfrac{{{a}_{8}}^{r+k}}{{{a}_{9}}^{r+k}} & {{\log }_{e}}{{a}_{9}}^{r}{{a}_{10}}^{k} \\ \end{matrix} \right|$
Now we know that $\dfrac{{{a}_{2}}}{{{a}_{1}}}=\dfrac{{{a}_{3}}}{{{a}_{2}}}=....=\dfrac{{{a}_{10}}}{{{a}_{9}}}=d$ hence we get.
$\left| \begin{matrix} {{\log }_{e}}{{\left( \dfrac{1}{d} \right)}^{r+k}} & {{\log }_{e}}{{\left( \dfrac{1}{d} \right)}^{r+k}} & {{\log }_{e}}{{a}_{3}}^{r}{{a}_{4}}^{k} \\ {{\log }_{e}}{{\left( \dfrac{1}{d} \right)}^{r+k}} & {{\log }_{e}}{{\left( \dfrac{1}{d} \right)}^{r+k}} & {{\log }_{e}}{{a}_{6}}^{r}{{a}_{7}}^{k} \\ {{\log }_{e}}{{\left( \dfrac{1}{d} \right)}^{r+k}} & {{\log }_{e}}{{\left( \dfrac{1}{d} \right)}^{r+k}} & {{\log }_{e}}{{a}_{9}}^{r}{{a}_{10}}^{k} \\ \end{matrix} \right|$
Now we know that if two rows or two columns of the determinant are the same then the value of the determinant is 0.
Hence we get $\left| \begin{matrix} {{\log }_{e}}{{\left( \dfrac{1}{d} \right)}^{r+k}} & {{\log }_{e}}{{\left( \dfrac{1}{d} \right)}^{r+k}} & {{\log }_{e}}{{a}_{3}}^{r}{{a}_{4}}^{k} \\ {{\log }_{e}}{{\left( \dfrac{1}{d} \right)}^{r+k}} & {{\log }_{e}}{{\left( \dfrac{1}{d} \right)}^{r+k}} & {{\log }_{e}}{{a}_{6}}^{r}{{a}_{7}}^{k} \\ {{\log }_{e}}{{\left( \dfrac{1}{d} \right)}^{r+k}} & {{\log }_{e}}{{\left( \dfrac{1}{d} \right)}^{r+k}} & {{\log }_{e}}{{a}_{9}}^{r}{{a}_{10}}^{k} \\ \end{matrix} \right|=0$ .
This means nothing but $\left| \begin{matrix} {{\log }_{e}}{{a}_{1}}^{r}{{a}_{2}}^{k} & {{\log }_{e}}{{a}_{2}}^{r}{{a}_{3}}^{k} & {{\log }_{e}}{{a}_{3}}^{r}{{a}_{4}}^{k} \\ {{\log }_{e}}{{a}_{4}}^{r}{{a}_{5}}^{k} & {{\log }_{e}}{{a}_{5}}^{r}{{a}_{6}}^{k} & {{\log }_{e}}{{a}_{6}}^{r}{{a}_{7}}^{k} \\ {{\log }_{e}}{{a}_{7}}^{r}{{a}_{8}}^{k} & {{\log }_{e}}{{a}_{8}}^{r}{{a}_{9}}^{k} & {{\log }_{e}}{{a}_{9}}^{r}{{a}_{10}}^{k} \\ \end{matrix} \right|=0$
Now this is always true. Which means the value of determinant is 0 independent of our choice of r and k.
Hence we can choose infinitely many r and k.
Now since we have S is a set of (r, k) such that $r,k\in N$ we can say that there are infinitely many elements in S.
Hence, option (a) is the correct option.

Note: Note that here by looking we get an idea that the determinant can be reduced to a simpler form. Here we can also use row transformation or some other column transformation to solve. Just make sure we get successive terms together to simplify further. Also, note that the value of the determinant is 0 always which was the given equation and hence we have an infinite solution and not 0 solutions. Also if we do not want to use the result that the value of the determinant is 0 when two rows and columns are the same we can use one more transformation in which we subtract the same rows or columns obtained, Hence we get one row or column as 0 which means the value of the determinant is 0.